package leetcode.editor.cn;

/**
* @Author: Dempsey
* @Data: 2021-05-08 13:14:47
*/

//给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。 
//
// 单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。 
//
// 
//
// 示例 1： 
//
// 
//输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "AB
//CCED"
//输出：true
// 
//
// 示例 2： 
//
// 
//输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SE
//E"
//输出：true
// 
//
// 示例 3： 
//
// 
//输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "AB
//CB"
//输出：false
// 
//
// 
//
// 提示： 
//
// 
// m == board.length 
// n = board[i].length 
// 1 <= m, n <= 6 
// 1 <= word.length <= 15 
// board 和 word 仅由大小写英文字母组成 
// 
//
// 
//
// 进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？ 
// Related Topics 数组 回溯算法 
// 👍 893 👎 0


public class P79{
	public static void main(String[] args) {
		Solution solution = new P79().new Solution();
	}
	
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
	public boolean exist(char[][] board, String word) {
		int h = board.length, w = board[0].length;
		boolean[][] visited = new boolean[h][w];
		for (int i = 0; i < h; i++) {
			for (int j = 0; j < w; j++) {
				if (dfs(board, visited, i, j, word, 0)) {
					return true;
				}
			}
		}
		return false;
	}

	public boolean dfs(char[][] board, boolean[][] visited, int i, int j, String s, int k) {
		if (board[i][j] != s.charAt(k)) {
			return false;
		}
		if (k == s.length() - 1) {
			return true;
		}
		visited[i][j] = true;
		int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
		boolean result = false;
		for (int[] dir : directions) {
			int newi = i + dir[0], newj = j + dir[1];
			if (newi >= 0 && newi < board.length && newj >= 0 && newj < board[0].length) {
				if (!visited[newi][newj]) {
					if (dfs(board, visited, newi, newj, s, k + 1)) {
						result = true;
						break;
					}
				}
			}
		}
		visited[i][j] = false;
		return result;
	}
}
//leetcode submit region end(Prohibit modification and deletion)

}